Since E[Y] = E[E[Y/X]], can i infer that Y = E[Y/X]? Therefore i conclude that:

Cov(X, E[Y/X])

= E[(X - E[X])(E[Y/X] - E[E[Y/X]])]

= E[(X - E[X])(Y- E[Y])]

= Cov(X, Y)

Thanks in advance!

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- Mar 28th 2009, 08:37 PMnoob mathematicianProving question
Since E[Y] = E[E[Y/X]], can i infer that Y = E[Y/X]? Therefore i conclude that:

Cov(X, E[Y/X])

= E[(X - E[X])(E[Y/X] - E[E[Y/X]])]

= E[(X - E[X])(Y- E[Y])]

= Cov(X, Y)

Thanks in advance! - Mar 29th 2009, 09:47 AMLaurent
No you can't. But, it is true that $\displaystyle E[f(X)E[Y|X]]=E[f(X)Y]$ for any function $\displaystyle f$ (that's a characterization of the conditional expectation).

Quote:

Therefore i conclude that:

Cov(X, E[Y/X])

= E[(X - E[X])(E[Y/X] - E[E[Y/X]])]

= E[(X - E[X])(Y- E[Y])]

= Cov(X, Y)

Thanks in advance!